Smallest possible uncertainty in the positon of an electron

1. The problem statement, all variables and given/known data
Show that the smallest possible uncertainty in the position of an electron whose speed is given by [tex]\beta=v/c[/tex] is
[tex]\Delta x_{min}=\frac{h}{4\pi m_{0}c}(1-\beta^{2})^{1/2}[/tex]

2. Relevant equations
[tex]\Delta x \Delta p=\frac{h}{4\pi}[/tex]

[tex]p=mv= \frac{m_{0}}{\sqrt{1-\beta^{2}}}v[/tex]

3. The attempt at a solution
so from the momentum equation, i multiply in c:

So then i assumed that for minimum [tex]\Delta x[/tex] we need maximum [tex]\Delta p[/tex] and thus maximum [tex]\Delta \beta[/tex], which will give us [tex]\Delta \beta=1[/tex] (because max value of v=c)

But then when i sub this into the uncertainty equation i still have the [tex](1+\frac{\beta^{2}}{1-\beta^{2}})[/tex] term i cant get rid off.

So then i assumed that for minimum [tex]\Delta x[/tex] we need maximum [tex]\Delta p[/tex] and thus maximum [tex]\Delta \beta[/tex], which will give us [tex]\Delta \beta=1[/tex] (because max value of v=c)

But then when i sub this into the uncertainty equation i still have the [tex](1+\frac{\beta^{2}}{1-\beta^{2}})[/tex] term i cant get rid off.

So using this method, the maximum [itex]\Delta p[/itex] would be [itex]m_{0}c(1-\beta^{2})^{-3/2}[/itex], but this is greater than [itex]m_{0}c(1-\beta^{2})^{-1/2}[/itex] by a factor of [itex](1-\beta^{2})^{-1}[/itex] and the energy argument provides tighter restrictions on [itex]\Delta p[/itex].

hi bina, I think you are close to the answear in the very first post...
just cancel the term [tex](1+\frac{\beta^{2}}{1-\beta^{2}})[/tex], since b should be regarded as a given value. it is only dv we need to deal with. however, dv is c*db right? :)